Impulse turbine optimum velocity ratio The hydraulic efficiency of an impulse turbine is maximum when the runner (wheel) peripheral speed is what fraction of the jet velocity?

Introduction / Context:Impulse turbines (e.g., Pelton) extract energy by changing the momentum of a high-speed jet. The wheel speed relative to jet speed controls how much of the jet's kinetic energy is converted to shaft work. There is a well-known optimum speed ratio that maximizes hydraulic efficiency under typical assumptions.

Given Data / Assumptions:

• Ideal impulse action with no shaft work from pressure changes (atmospheric pressure at runner).
• Jet deflection approximates 180° minus losses; relative exit speed nearly equals relative inlet speed.
• Neglecting mechanical losses for the hydraulic-optimum reasoning.

Concept / Approach:The hydraulic power depends on the change in whirl component of velocity across the runner. Treating losses modestly, differentiating power with respect to runner speed yields a maximum when the wheel speed equals approximately half the jet velocity.

Step-by-Step Solution:Let jet speed be V and wheel peripheral speed be u.Hydraulic efficiency depends on the product of mass flow and change in whirl velocity (V − u) terms.Optimization of the idealized expression gives u/V ≈ 0.5.Therefore, the best hydraulic efficiency is achieved when u is one-half of V.

Verification / Alternative check:Pelton design guides and textbook derivations converge on the 0.46–0.5 practical range; ideal analysis gives 0.5, with real losses shifting slightly lower.

Why Other Options Are Wrong:One-fourth or three-fourth deviate from the optimum, reducing momentum change effectiveness. “Double” or “equal to jet” produce very low relative jet deflection or zero energy extraction.

Common Pitfalls:Confusing mechanical (overall) efficiency with hydraulic efficiency; ignoring the impact of bucket losses that slightly reduce the optimal ratio in practice.

Final Answer:one-half